${\text{AlF}}_3$ is soluble in $\text{HF}$ only in presence of $\text{KF}$. It is due to the formation of:

${\text{AlF}}_3$ is insoluble in $\text{HF}$ because the ${\text{F}}^{-}$ ion is not available (it is tied up in ${\text{HF}}_2^{-}$). However, in the presence of $\text{KF}$, a soluble complex is formed:
$$3\text{KF}+{\text{AlF}}_3\to {\text{K}}_3[{\text{AlF}}_6]$$
This complex is an ionic compound which dissolves in the medium.